Consider the titration of 30.0 mL of 0.251 M weak base B (Kb = 1.3 x 10⁻¹⁰) with 0.150 M HI. How many mL of the HI would be required to reach the halfway point?

Answer 1

B + HI ==> BH⁺ + I^-

moles B present = 30.0 ml x 1 L / 1000 ml x 0.251 mol/L = 7.53x10^-3 moles

To reach halfway of neutralization would require 1/2 x 7.53x10^-3 moles = 3.77x10^-3 moles

3.77x10^-3 moles x 1 L / 0.150 mols = 0.0251 L x 1000 ml / L = 25.1 mls of HI